// Part 1 about Regular Expression Matching//==========================================object CW8a {abstract class Rexpcase object ZERO extends Rexpcase object ONE extends Rexpcase class CHAR(c: Char) extends Rexpcase class ALT(r1: Rexp, r2: Rexp) extends Rexp case class SEQ(r1: Rexp, r2: Rexp) extends Rexp case class STAR(r: Rexp) extends Rexp // some convenience for typing in regular expressionsimport scala.language.implicitConversions import scala.language.reflectiveCalls def charlist2rexp(s: List[Char]): Rexp = s match { case Nil => ONE case c::Nil => CHAR(c) case c::s => SEQ(CHAR(c), charlist2rexp(s))}implicit def string2rexp(s: String): Rexp = charlist2rexp(s.toList)implicit def RexpOps (r: Rexp) = new { def | (s: Rexp) = ALT(r, s) def % = STAR(r) def ~ (s: Rexp) = SEQ(r, s)}implicit def stringOps (s: String) = new { def | (r: Rexp) = ALT(s, r) def | (r: String) = ALT(s, r) def % = STAR(s) def ~ (r: Rexp) = SEQ(s, r) def ~ (r: String) = SEQ(s, r)}// (1a) Complete the function nullable according to// the definition given in the coursework; this // function checks whether a regular expression// can match the empty stringdef nullable (r: Rexp) : Boolean = r match { case ZERO => false case ONE => true case CHAR(_) => false case ALT(r1, r2) => nullable(r1) || nullable(r2) case SEQ(r1, r2) => nullable(r1) && nullable(r2) case STAR(_) => true}// (1b) Complete the function der according to// the definition given in the coursework; this// function calculates the derivative of a // regular expression w.r.t. a characterdef der (c: Char, r: Rexp) : Rexp = r match { case ZERO => ZERO case ONE => ZERO case CHAR(d) => if (c == d) ONE else ZERO case ALT(r1, r2) => ALT(der(c, r1), der(c, r2)) case SEQ(r1, r2) => if (nullable(r1)) ALT(SEQ(der(c, r1), r2), der(c, r2)) else SEQ(der(c, r1), r2) case STAR(r1) => SEQ(der(c, r1), STAR(r1))}// (1c) Complete the function der according to// the specification given in the coursework; this// function simplifies a regular expression;// however it does not simplify inside STAR-regular// expressionsdef simp(r: Rexp) : Rexp = r match { case ALT(r1, r2) => (simp(r1), simp(r2)) match { case (ZERO, r2s) => r2s case (r1s, ZERO) => r1s case (r1s, r2s) => if (r1s == r2s) r1s else ALT (r1s, r2s) } case SEQ(r1, r2) => (simp(r1), simp(r2)) match { case (ZERO, _) => ZERO case (_, ZERO) => ZERO case (ONE, r2s) => r2s case (r1s, ONE) => r1s case (r1s, r2s) => SEQ(r1s, r2s) } case r => r}// (1d) Complete the two functions below; the first // calculates the derivative w.r.t. a string; the second// is the regular expression matcher taking a regular// expression and a string and checks whether the// string matches the regular expressiondef ders (s: List[Char], r: Rexp) : Rexp = s match { case Nil => r case c::s => ders(s, simp(der(c, r)))}// main matcher functiondef matcher(r: Rexp, s: String): Boolean = nullable(ders(s.toList, r))// (1e) Complete the size function for regular// expressions according to the specification // given in the coursework.def size(r: Rexp): Int = r match { case ZERO => 1 case ONE => 1 case CHAR(_) => 1 case ALT(r1, r2) => 1 + size(r1) + size (r2) case SEQ(r1, r2) => 1 + size(r1) + size (r2) case STAR(r1) => 1 + size(r1)}// some testing data/*matcher(("a" ~ "b") ~ "c", "abc") // => truematcher(("a" ~ "b") ~ "c", "ab") // => false// the supposedly 'evil' regular expression (a*)* bval EVIL = SEQ(STAR(STAR(CHAR('a'))), CHAR('b'))matcher(EVIL, "a" * 1000 ++ "b") // => truematcher(EVIL, "a" * 1000) // => false// size without simplificationssize(der('a', der('a', EVIL))) // => 28size(der('a', der('a', der('a', EVIL)))) // => 58// size with simplificationsize(simp(der('a', der('a', EVIL)))) // => 8size(simp(der('a', der('a', der('a', EVIL))))) // => 8// Java needs around 30 seconds for matching 28 a's with EVIL. //// Lets see how long it takes to match strings with // 0.5 Million a's...it should be in the range of some// seconds.def time_needed[T](i: Int, code: => T) = { val start = System.nanoTime() for (j <- 1 to i) code val end = System.nanoTime() (end - start)/(i * 1.0e9)}for (i <- 0 to 5000000 by 500000) { println(i + " " + "%.5f".format(time_needed(2, matcher(EVIL, "a" * i))))}*/}